Assuming a 2D boundary layer on a flat surface with no pressure gradient (i.e. $dP/dx=0$), suppose the $x$-component of velocity, $U$, has a profile $$\dfrac U{U_e}=\dfrac {2y}\delta -\dfrac{y^2}{\delta^2}$$ where $\delta$ is the boundary layer thickness and $U_e$ is the free-stream velocity.

It can be made non-dimensional, so, $$\dfrac U{U_e} = 2\eta-\eta^2,\qquad\eta=\dfrac y\delta$$

This profile satisfies all the needed boundary conditions for shear, penetration, and slipping. Using the continuity equation $\nabla\cdot \vec v=0$, one can find that$$V=-\int \dfrac{\partial U}{\partial x}dy+f(x)$$and reason that $f=0$ to satisfy boundary conditions. When evaluating all this, I get $$\dfrac V{U_e} = \left( \eta^2 - \dfrac23 \eta^3 \right) \dfrac{d\delta}{dx}$$

which makes sense so far. If I try to now solve for an expression for $\delta$ by using the Momentum Integral Equation, I end up with an ODE with $x$ only, and everything works out.

However, my problem arises when trying to satisfy Prandtl's boundary layer equation, namely,

$$U \dfrac{\partial U}{\partial x}+V\dfrac{\partial U}{\partial y}=\nu \dfrac{\partial^2 U}{\partial y^2}$$

Upon substituting all the variables, I end up with an equation of the form

$$g(\eta) \dfrac {d\delta}{dx} = \dfrac{\alpha}{\delta}$$ where $\alpha$ is a constant and $g(\eta)$ is a fourth-order polynomial function of $\eta$. This is very confusing to me, because $\delta$ should be a function of $x$ only, and $x$ is independent of $y$.

Why do we not end up with a separable PDE to allow for a direct solution of $\delta(x)$?


1 Answer 1


Edit - The reason why the velocity field with $u(\eta) = 2 \eta - \eta^2$ is not part of a solution of Prandtl equation for the boundary layer. The reason why $u(\eta) = 2\eta-\eta^2$ and the $v$-component derived from integration of the incompressibility constraint don't satisfy Prandtl equation for the boundary layer is that this is not a solution of the equations, but $u(\eta)$ is only an approximation of an exact solution of the equations.

An exact solution of the laminar boundary layer is Blasius solution, that you can find there https://gitlab.com/davideMontagnani/fluidmechanics-ita/-/blob/master/template/mainEse.pdf?ref_type=heads. Unfortunately, Blasius solution doesn't have a simple closed expression (and this is the reason why sometimes approximate velocity profiles are used), but you can prove that the velocity field satisfies Prandtl equations.

Old answer.

(It's quite late here, so some algebra mistake could occur, but the overall answer should be good enough. I'll check it tomorrow morning)

Let's split this answer in some steps, namely:

  • integral thickness of the boundary layer
  • Von Karman integral equation for the boundary layer
  • Von Karman integral equation for the laminar boundary layer, with the proper expression of the wall stress, and assumption of no pressure gradient along the wall (and thus, constant asymptotic velocity, $U(x) = \overline{U}$

Integral thickness

Some integral thicknesses of the boundary layers can be defined, related with defect of mass and momentum flux (if compared with the inviscid flow), that appear in VK integral equation of the boundary layer $$ \begin{aligned} \delta_1(x) & := \int_{y=0}^{\infty} \left(1 - \frac{u(x,y)}{U(x)} \right) dy \\ \delta_2(x) & := \int_{y=0}^{\infty} \frac{u(x,y)}{U(x)} \left(1 - \frac{u(x,y)}{U(x)} \right) dy \\ \end{aligned} $$ and the shape ratio as $$H := \frac{\delta_1(x)}{\delta_2(x)} \ ,$$ that is (at least analytically) independent on the coordinate $x$.

Von Karman integral equation of the boundary layer

VK integral equation of the boundary layer reads (derivation in the link in the comment to your question)

$$\dfrac{d \theta}{d x} + (2+H)\frac{\theta}{U}\frac{dU}{dx} = \frac{c_f}{2} \ ,$$

being $c_f(x) = \frac{\tau_w(x)}{\frac{1}{2}\rho U(x)^2}$ the friction coefficient at wall.

Von Karman integral and laminar boundary layer of the question

For a laminar boundary layer, with no pressure gradient along the wall ($0 = \partial_x P = \frac{dU}{dx}$), with the velocity profile of your question, the momentum thickness is $$\theta(x) = \delta(x) \frac{2}{15}$$ and the friction coefficient is $$\tau_w(x) = \mu \frac{\partial u}{\partial y}\bigg|_{y=0} = 2 \frac{\mu U(x)}{\delta}$$ so that the VK integral equation becomes

$$\frac{2}{15} \frac{d \delta(x)}{dx} = \frac{\nu}{U \delta(x)} \ ,$$

whose solution reads

$$\delta^2(x) - \delta^2_0 = 15 \frac{\nu \left(x-x_0\right)}{U} \ ,$$

so that if we prescribe $\delta(x_0 = 0) = 0$ we get the power $1/2$ thickness of the laminar boundary layer as

$$ \delta(x) = \sqrt{15\frac{\nu x}{U}} \ .$$

  • $\begingroup$ There;s a missing factor of 1/2 in the step where you integrate, but I did get that part. My general question was why the solution we get here does not satisfy the original x-momentum equation. If we substitute in $\delta$, $V$, $U$, and corresponding derivatives, the x-momentum equation is not satisfied. The best answer i came to find was that the velocity profile that we start with is an approximate solution and thus wont satisfy the momentum exactly. $\endgroup$
    – user256872
    Feb 9 at 1:42
  • $\begingroup$ Especially given how the VK equation is derived -- it follows directly from the x-momentum equation by integrating from 0 to $\infty$, and no approximations were made during those steps. $\endgroup$
    – user256872
    Feb 9 at 1:47
  • $\begingroup$ This solution needs to satisfy Prandtl equations of the boundary layers, not the full Navier--Stokes equations. We're solving only the "inner flow" (the boundary layer close to the wall), while the outer flow is given by the inviscid flow solution of Euler equations. I'll edit my answer to add this remark $\endgroup$
    – basics
    Feb 9 at 6:35
  • $\begingroup$ anyways. If you want a solution of the laminar boundary layer that has the $u$ and $v$-components satisfying Prandtl equations of the boundary layer, you need an exact solution of those equations, like Blasius solution (see the ref.), and not approximate solutions like the $u$-component profile given in your question $\endgroup$
    – basics
    Feb 9 at 7:30

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